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Director
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A certin deck of cards contains 2 blue cards,2 red cards,2 [#permalink]
 06 Jan 2004, 12:53
** A certin deck of cards contains 2 blue cards,2 red cards,2 yellow cards and 2 green cards.If two cards are randomly drawn from the deck,what is the probability that they both will not be blue?
1) 15/28
2) 1/4
3) 9/16
4) 1/32
5) 1/16
** A bag of 10 marbles contains 3 red marbles and 7 blue marbles.If 2 marbles are selected at randon ,what is the probability that at least one marble is blue?
1) 21/50
2) 3/13
3) 47/50
4) 14/15
5) 1/5
Please provide an explantion.
Vivek.
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"Start By Doing What Is Necessary ,Then What Is Possible & Suddenly You Will Realise That You Are Doing The Impossible"
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6/8 * 5/7 = 15/28
1- (Chance of red, red)
1- (3/10 * 2/9)
14/15
Will post reasoning later, but it ought to be pretty obvious...
(post edited for typographical error)
Last edited by stoolfi on 06 Jan 2004, 14:08, edited 1 time in total.
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Director
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stoolfi wrote: 6/8 * 5/7 = 15/28
1- (Chance of red, red)
1- (3/10 * 2/9)
13/15
Will post reasoning later, but it ought to be pretty obvious...
stoolfi - i'm sure you mean't 14/15. you're becoming like me. 
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Intern
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I agree with the first one. But wouldn't 2nd one be 14/15?
P of both red is 3/10 * 2*9 =6/90 = 1/15.
P of at least one blue is 1 - 1/15 = 14/15.
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Director
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drodgers wrote: I agree with the first one. But wouldn't 2nd one be 14/15?
P of both red is 3/10 * 2*9 =6/90 = 1/15.
P of at least one blue is 1 - 1/15 = 14/15.
yes - that was a titleistesque stoolfi typo. i give him credit!
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Director
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stoolfi wrote: 6/8 * 5/7 = 15/28
1- (Chance of red, red)
1- (3/10 * 2/9)
14/15
Will post reasoning later, but it ought to be pretty obvious...
(post edited for typographical error)
This is how I had done it...
Both will not be blue= 1 - both of them are blue
= 1 - (2/8 * 1/7)
= 1 - (1/28)
= 15/28
Can anybody tell me where I have gone wrong.
I agree with the second answer.
Vivek.
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"Start By Doing What Is Necessary ,Then What Is Possible & Suddenly You Will Realise That You Are Doing The Impossible"
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Director
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vivek_dj wrote: stoolfi wrote: 6/8 * 5/7 = 15/28
1- (Chance of red, red)
1- (3/10 * 2/9)
14/15
Will post reasoning later, but it ought to be pretty obvious...
(post edited for typographical error) This is how I had done it... Both will not be blue= 1 - both of them are blue = 1 - (2/8 * 1/7) = 1 - (1/28) = 15/28 Can anybody tell me where I have gone wrong. I agree with the second answer. Vivek. Probability of any color except blue (1) * probability of any color except blue (2)
6/8*5/7 =30/56 = 15/28
You just got there differently that's all
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Director
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Sorry Titleist ..there was typo in my last post...
The answer I would get from my approach is 27/28.
I comprehend the solution given by you and stoolfi.
I just wanna know where am I going wrong in this approach.
Regards,
Vivek.
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"Start By Doing What Is Necessary ,Then What Is Possible & Suddenly You Will Realise That You Are Doing The Impossible"
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Director
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vivek, I see your point quite clearly. your answer is right since the question does specifically ask the probability of not getting two blue balls. since there's no answer choice for 27/28 I'm assuming that it's asking: "what's the probability that there are no blue balls chosen" where did you find this problem?
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Re: Probability Q's [#permalink]
07 Jan 2004, 04:15
Quote: A certin deck of cards contains 2 blue cards,2 red cards,2 yellow cards and 2 green cards.If two cards are randomly drawn from the deck,what is the probability that they both will not be blue?
1) 15/28
2) 1/4
3) 9/16
4) 1/32
5) 1/16 6C2/8C2 = 15/28 Quote: A bag of 10 marbles contains 3 red marbles and 7 blue marbles.If 2 marbles are selected at randon ,what is the probability that at least one marble is blue?
1) 21/50
2) 3/13
3) 47/50
4) 14/15
5) 1/5
Please provide an explantion.
1- 3C2 / 10C2 = 14/15
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Re: Probability Q's
[#permalink]
07 Jan 2004, 04:15
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